Applications
of Vectors
Definition:
Resultant: The result of two vectors acting on a point at the same
time.
Equilibrant: The opposite vector of the resultant.
Find the resultant by ADDING vectors.
Two Methods to solve Application Problems.
A). Resolve the vectors into component form. Then add.
Two forces act on a point.
Vector A has direction angle = 30o and magnitude = 50
Vector B has direction angle = 135o and magnitude = 20
1. Write each vector in Component form – start with trig form
2. Add the components to find the Resultant Vector
3. Find the magnitude and Amplitude of the Resultant
Two forces act on a point.
Vector A has direction angle = 30o and magnitude = 50
Vector B has direction angle = 135o and magnitude = 20
𝑥 = 50 cos 30° = 25 3
𝑦 = 50 sin 30° = 25
25 3, 25
𝑥 = 20 cos 135° = −10 2
𝑦 = 20 sin 135° = 10 2
−10 2, 10 2
25 3 + −10 2 , 25 + 10 2
25 3 − 10 2, 25 + 10 2
resultant
Mag=
29.159,39.142
48.809
amp = 53.316°
𝐵
𝐴
Two Methods to solve Application Problems.
A). Resolve the vectors into component form. Then add.
Two forces act on a point . Given the angle between the angles
Vector A has magnitude = 15
Vector B has magnitude = 23
The angle between the two angles is 70o.
Make one the x-axis
15 cos 0°, 15 sin 0°
15,0
Resultant has a magnitude of
23 cos 70°, 23 sin 70°
7.866,21.613
31.464
and amplitude of 43.386°
Two Method to solve Application Problems.
B). Parallelogram Method - Law of Cosines.
Two forces act on a point . Given the angle between the angles
Vector A has magnitude = 15
Vector B has magnitude = 22
The angle between the two angles is 100o.
1) Draw a Parallelogram.
2) Find the angle opposite the resultant – using the Law of Cosines
3) Use the Law of Sines to find the angles.
Two forces act on a point . Given the angle between the angles
Vector A has magnitude = 15
Vector B has magnitude = 22
The angle between the two angles is 100o.
1) Draw a Parallelogram.
2) Find the angle opposite the resultant – using the Law of Cosines
3) Use the Law of Sines to find the angles.
𝑐 2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶
100⁰
𝑐 2 = 152 + 222 − 2 15 22 cos 80°
𝑐 = 225 + 484 − 660 ∗ .1736481777
𝑐 =24.380
x
15
sin 80
sin 𝑥
=
24.380
22
𝑥 = 62.707°
Two Method to solve Application Problems.
B). Parallelogram Method - Law of Cosines.
Two forces at on a point.
Vector A has direction angle = 75o and magnitude = 45
Vector B has direction angle = 310o and magnitude = 27
Two forces act on a point.
Vector A has direction angle = 75o and magnitude = 45
Vector B has direction angle = 310o and magnitude = 27
𝑥 = 45 cos 75
𝑥 = 27 cos 310
𝑦 = 45 sin 75
𝑦 = 27 sin 310
11.647,43.467
17.355, −20.683
29.002,22.784
36.881
38.153°
x
𝑐2
=
𝑐=
452
+ 272
y
− 2 45 27 cos 55
452 + 272 − 2 45 27 cos 55
𝑐 = 36.881
sin 55
sin 𝑦
=
36.881
45
sin 𝑦 = .999
sin−1 .999 = 88.153°
𝑥 = 38.153°
Assignment:
Applications 1: Methods